Question

Solve the following inequality:
$\sin x⩽\frac{1}{2}.$

Answer 1

$\sin x\le \frac{1}{2}$

we can say that

$xϵR-(n\pi +\frac{\pi }{6},(n+1)\pi -\frac{\pi }{6})$

i.e, $xϵR-(n\pi +\frac{\pi }{6},(n+1)\pi -\frac{\pi }{6})$